Energy `E` of a hydrogen atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) eV`. The energy of a photon ejected when the electron jumps from `n = 3` state to `n = 2` state of hydrogen is approximately

Options

A. `1.5eV`
B. `0.85eV`
C. `3.4eV`
D. `1.9eV`

Answer

D

Explanation

`E=-(13.6)/(n^2)Ev`
the energy of photon
`=E_3-E_2=(-13.6)/(3)^2+(13.6)/(2)^2=13.6[(-1)/(9)+(1)/(4)]`
` =13.6[(-4+9)/(36)]=13.6xx(5)/(36)=1.9eV`.